In triangle $ABC$, the angle bisectors are $AD$, $BE$, and $CF$, which intersect at the incenter $I$.  If $\angle ACB = 38^\circ$, then find the measure of $\angle AIE$, in degrees.
Explanation: Since $AD$ is an angle bisector, $\angle BAI = \angle BAC/2$.  Since $BE$ is an angle bisector, $\angle ABI = \angle ABC/2$.  As an angle that is external to triangle $ABI$, $\angle AIE = \angle BAI + \angle ABI = \angle BAC/2 + \angle ABC/2$.

[asy]
import geometry;

unitsize(0.3 cm);

pair A, B, C, D, E, F, I;

A = (2,12);
B = (0,0);
C = (14,0);
I = incenter(A,B,C);
D = extension(A,I,B,C);
E = extension(B,I,C,A);
F = extension(C,I,A,B);

draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$I$", I, NNE);
[/asy]

Since $\angle ACB = 38^\circ$, \[\angle AIE = \frac{\angle BAC + \angle ABC}{2} = \frac{180^\circ - \angle ACB}{2} = \frac{180^\circ - 38^\circ}{2} = \boxed{71^\circ}.\]